Application of the Fundamental Theorem of Calculus (cosmological red-shift)

[binbagsss]

Homework Statement

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I am stuck on the section of my lecture notes attached, where it says that equation 4.20 follows from 4.18 via an application of the fundamental theorem of calculus

Homework Equations

FoC:

if $f$ is cts on $[a,b]$ then the function :

$F(x)=\int\limits^{x}_{c} f(t) dt$

has a derivaitve at every point in $[a,b]$ and

$\frac{dF}{dx}=\frac{d}{dx} \int\limits^{x}_{a} f(t) dt=f(x)$

The Attempt at a Solution

[/B]
I can't see the link since 4.18 says that:

Let me call $\int dt \frac{1}{a(t)} = b(t)$, 4.18 says that $b(t_2)-b(t_1)=r_1$ whilst 4.20 is looking at something completely different : $b(t_2 + \Delta t_2)-b(t_1+\Delta t_1)$(and so rather than an application of the FoC I thought it was expansion of $b(t)$ assuming $\Delta t_2$ is small)

So it looks like we have used 4.18 s.t the RHS can be set to zero if there is some property from the FoC that allows to do some sort of split on the limits, a corollary following from the FoC or something that I'm not seeing?

Many thanks in advance.

 

[kdv]

Homework Statement

View attachment 221325
[/B]
I am stuck on the section of my lecture notes attached, where it says that equation 4.20 follows from 4.18 via an application of the fundamental theorem of calculus

Homework Equations

FoC:

if $f$ is cts on $[a,b]$ then the function :

$F(x)=\int\limits^{x}_{c} f(t) dt$

has a derivaitve at every point in $[a,b]$ and

$\frac{dF}{dx}=\frac{d}{dx} \int\limits^{x}_{a} f(t) dt=f(x)$

The Attempt at a Solution

[/B]
I can't see the link since 4.18 says that:

Let me call $\int dt \frac{1}{a(t)} = b(t)$, 4.18 says that $b(t_2)-b(t_1)=r_1$ whilst 4.20 is looking at something completely different : $b(t_2 + \Delta t_2)-b(t_1+\Delta t_1)$(and so rather than an application of the FoC I thought it was expansion of $b(t)$ assuming $\Delta t_2$ is small)
.

Exactly. Just do a Taylor expansion, for example
$b(t_2 + \Delta t_2) \approx b(t_2) + \Delta t_2 \frac{db}{dt}(t_2)$

The second term is simply $\Delta t_2 ~ \frac{1}{a(t_2)}$.